Download Mechanics Aide Memoire

Mechanics 2 Aide Memoire
Work done – measured in Joules
Energy lost due to overcoming a force (no work done if force acts perpendicular to
the direction of motion)
Work done =
 Fdx
or Fx for a constant force, F over a distance, x.
= Area under a force/distance graph
=
 Pdt
or Pt for constant power, P over t seconds
= change in kinetic energy
Power – measured in Watts
The rate at which work is done
Power = Force x Velocity
= d (work done)
dt
IMPORTANT – At maximum or constant velocity, no acceleration so forces are
balanced.
Example 1
Solution
a) a = dv = -4e2t
dt
2et
Using F = ma gives
b) i) F.v = -16e2t .
8et
F = 4 -4e2t
2et
= -16e2t
8et
1 - 2e2t = -16e2t + 32e2te2t + 16etet
2et
= -16e2t + 32e4t + 16e2t
= 32e4t
ii) F.v = Power
Work done = ∫Power dt
=
∫32e
4t
= 8e4t
= 648 – 8
= 640J
Example 2
dt
Solution
a)
Tractive
Force, T
R
50
resistive
forces
1960N
1200g
b) i) Constant velocity so no acceleration i.e. forces are balanced
T = R + 1200gsin5o
Resolving along the plane
T = 1960 + 11760sin5o
T = 2985 N ( 4 sig. figs.)
ii)
Rate of doing work = Power
= Force x Velocity
= 2985 x 15
(force used is tractive force as we want
the power of the engine)
= 44775 W
c) At maximum speed, no acceleration i.e. forces are balanced so T = 2985 N
Power = Force x Velocity
60000 = 2985v
v = 20.1 ms-1
Conservation of Energy – energy is conserved unless work is done. In these cases
energy lost = work done
There are 3 main types of energy at this level
i)
ii)
iii)
Potential energy (P.E.) – mass x acceleration x height
Kinetic energy (K.E.) - ½ x mass x (velocity) 2
Elastic Potential Energy (E.P.E) - ½ x spring constant x (extension) 2
N.B. spring constant, k = modulus, λ / natural length, l
λ measured in N; k measured in Nm-1
Example
Solution
a) E.P.E. = ½ x 50 x 0.032 = 0.0225 J
b) When spring reaches natural length (3cm) from start, no EPE (just KE and PE)
Conservation of Energy gives:
0.0225 = ½ x 0.02 x V2 + 0.02 x 9.8 x 0.03
V2 = 1.662
V = 1.29 ms-1 (3 s. f.)
c) Rocket comes to rest at highest point. Stationary so no KE, spring is not stretched
or compressed so no EPE.
All energy is PE
Conservation of Energy gives:
0.0225 = 0.02 x 9.8 x h
h = 11.5 cm (3 s. f.)
(h = height above start position)
Centre of Mass – the place where all the mass of an object appears to act.
A uniform lamina has an equal mass per unit area throughout the lamina.
Basically nothing more complicated than finding the ‘resultant moment’.
For a uniform shape, the centre of mass lies on the line(s) of symmetry.
If a shape is freely suspended, the centre of mass will always be directly below the
point it is suspended from.
Example
Solution
a) By symmetry, centre of mass is 5 cm from PQ.
b) Taking moments from PS, moments due to 2 masses and plate must be equivalent
to the moment due to the total mass acting at the centre of mass, a distance x from PS.
6 x m + 6 x m + 3 x 1 = (2m+1)x
12m + 3 = (2m+1)x
x = 12m+3
2m+1
(as required)
c) When freely suspended, centre of mass will be directly below the suspension
point.
12m+3
2m+1
centre of
mass
5cm
tan 450 = 5 x (2m+1)
12m+3
12m +3 = 10m +5
2m = 2
m = 1 kg
Example
Solution
a) Centre of mass must be on AP because it is the line of symmetry of a uniform
shape.
b)
B
Mass is determined by area
160g
4
256g
96g
Centre of mass
x
14
A
Turning force due to centre of mass = turning forces due to individual centres of mass
Using moments
160g x 4 + 96g x 14 = 256gx
x = 7.75cm
c) Lamina is freely suspended, so centre of mass will be below B.
B
7.75cm
θ
Centre of
mass
tan θ = 12.25
7.75
so θ = 57.7° (to 3 sig. fig.)
12.25 cm
Horizontal circular motion
Any particle moving in a circular orbit has a constantly changing velocity (direction it
is moving in changing). This means that there is an acceleration and therefore a
resultant force. The acceleration and resultant force are directed towards the centre of
the circular path.
v = rω
‘v’ is velocity, ‘r’ is radius, ‘ω’ is angular speed (rad s-1)
acceleration, a = v2
r
resultant force = ma = mv2
r
= rω2
= mrω2 (directed towards the centre of the circle)
IMPORTANT FACT – the resultant force is not an additional force, it is the resultant
of all the forces present.
Example
Solution
a)
θ
T
0.1g
b) Resolve vertically
Particle is not moving vertically so:
0.1g = Tcosθ
and cosθ = 0.8
so T = 0.1g = 1.225 N
0.8
(as required)
c) Resolve horizontally (should be a resultant force towards the centre)
Tsinθ = mrω2
sinθ = r
0.5
so 1.225 x r = 0.1 x r ω2
0.5
ω2 = 24.5
ω = 4.95 rad s-1 (3 s. f.)
Example
Solution
a) One revolution = 2π radians
so 8 revolutions per minute = 16π radians per minute
16π radians per minute = 16π radians per second
60
= 4π radians per second
15
b) Using v = rω
(as required)
r=v
ω
r = 0.75 x 15 = 0.8952….. = 0.90 metres (2 sig. fig.)
4π
c) Using a = v2
r
( or a = rω2 ) gives a = 0.63 ms-2 ( 2 sig. fig.)
directed towards the centre of the roundabout
Vertical Circular Motion
The major difference in the mechanics of vertical and circular motion is that in
vertical circular motion, the resultant force towards the centre (and hence the velocity
and acceleration) are constantly changing. However if we look at ‘snapshots’ of the
motion all the principles and equations relating to horizontal circular motion will still
be valid.
Particle is at its fastest at the bottom of its circular path and slowest at the top.
Vertical circular motion questions often involve a number of other principles such as
conservation of energy. They are usually worth a large number of marks.
Example
Solution
a) Draw diagram
D
60
r
R
mg
This is an energy question. We know the speed at B and that the track is smooth (no
work done against friction) so energy is conserved.
Energy at B (all K.E.) = ½ mx 7gr = 7mgr Joules
2
4
vertical height of D above B
Energy at D (mixture of K.E. and P.E.) = 1 mv2 + mg(r + rcos60)
2
Using C. of E.
7mgr
4
= 1 mv2 + mg(r + rcos60)
2
7gr = 2v2 +4g(r + rcos60)
remember cos60 = ½
7gr = 2v2 + 12gr = 2v2 + 6gr
2
2v2 = gr
v=
gr
2
b) Now for the vertical circular motion part.
We need to resolve towards the centre and show that the Normal Reaction force is
zero (i.e. no contact between skater and track).
Resolving towards O gives:
R + mg cos60 = mv2
r
(just like horizontal circular motion)
but
v=
gr
2
so R + mg = mgr
2
2r
R + mg = mg
2
2
R=0
(as required)
Example
Answer
a) Using Conservation of Energy
Energy at top (K.E. and P.E.) = 1m x 1rg + mrg
2
4
= 9mrg
8
vertical height above base
Energy ‘on side’ (K.E. And P.E.) = 1mv2 + mg rcosθ
2
Equating gives: 9mrg = 1mv2 + mgrcosθ
8
2
v2 = 9rg – 2grcosθ
4
v2 = 1rg (9 – 8cosθ)
4
(as required)
b) i) Diagram
R
v2 = 1rg (9 – 8cosθ)
4
θ
mg
Circular motion question – resolve towards the centre
mgcosθ - R = mv2
r
mgcosθ - R = mrg(9 – 8cosθ)
4r
R = mgcosθ - mg(9 – 8cosθ)
4
R = 3mgcosθ – 9mg
4
R = 3mg(cosθ - 3)
4
ii) when cosθ = 4
5
R = 3mg ( 4 - 3 )
5 4
R = 3mg
20
R is positive so particle still in contact
with hemisphere.
iii) When P loses contact with hemisphere there will be no reaction force i.e. R = 0
0 = 3mg(cosθ – 3)
4
cosθ = 3
4
θ = 41.4º (3 sig. fig.)
iv)
Using v2 = 1rg (9 – 8cosθ)
4
and
cosθ = 3
4
v2 = 1rg (9 – 8x3)
4
4
v2 = 3rg
4
so v =
3rg
2
(as required)